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I am playing around with a toy model of a transition metal complex where the HOMO are $d$-electron states of predominantly transition metal character. Let's say this is a $d^1$ (or $d^9$) system and the states are split in energy by ligand field effects. The relevant HOMO orbital is some linear combination of $d$-orbitals that correctly respects the symmetry of the transition metal complex.

I want to estimate the transition probability for a quadrupole $d$-to-$d$ transition from HOMO state. In other words, calculate $\langle f\vert r_i r_j \vert i \rangle$, where $\vert i \rangle$ and $\vert f \rangle$ are some $d$-orbitals I care about.

I can roughly estimate this transition amplitude by just using the Hydrogen $d$-orbital wavefunctions with an effective Bohr radius $a_{eff}$. My question is: what effective Bohr radius should I use?

For example, Copper is said to have a 3d-orbital radius of $\sim 0.6$ Angstrom (which is not the same as $a_{eff}$ I believe), but using the Slater estimate of the effective nuclear charge, I would guess $a_{eff} \approx a_0/Z_{eff} = a_0/13.2 \approx0.04$ Angstrom.

What effective Bohr radius should I actually use for very weakly hybridized transition metal orbitals?

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If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually

$$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$

or 32.4 pm.

As for Slaters approach, it is not enough to simply divide $a_0$ by the effective nuclear charge. You can only do this for an unshielded hydrogenic 1s orbital, otherwise it is generally necessary to account for the principal quantum number (in Slater's method, this may be an effective value not equal to the principal quantum number, see Slater's original article$^2$). The location of the maximum follows from the functional form of a Slater wave function:

$$\psi \propto r^{n^*-1}\exp\left({\frac{Z_{\textrm{eff}}}{n^*a_0}r}\right)$$

In general, if you change the nuclear charge of a hydrogenic atom you can perform the transformation $a_0 \rightarrow a_0/Z_{\textrm{eff}}$ to obtain the new orbitals$^3$.

The effective quantum number is n*=3 for a 3rd shell electron so that

$$r_{\textrm{Slater}} = \frac{n^{*2}}{Z_{\textrm{eff}}}=\frac{9}{13.2}\times0.529 Å = 0.361 Å $$

or 36.1 pm.

The difference in $r_{\textrm{max}}$ obtained by the two approaches is therefore not that dramatically different.

Of course there are two other things that seem to be important here: (1) whether using an accurate value of $r_{\textrm{max}}$ can be expected to provide you with an accurate value of the property you seek and (2) how to obtain an accurate estimate of $r_{\textrm{max}}$for a d-orbital of Cu in the transition metal complex.

References

  1. J.B. Mann, Atomic Structure Calculations II. Hartree-Fock wave functions and radial expectation values: hydrogen to lawrencium, LA-3691, Los Alamos Scientific Laboratory, USA, 1968.

  2. J.C. Slater. Atomic Shielding Constants. Phys. Rev., 36, 57, 1930.

  3. W. Greiner. Quantum Mechanics: An Introduction. Springer 1994.

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  • $\begingroup$ So if I want to get the actual estimated wavefunction, I would need to choose $a_0^*$ such that $\sqrt{\langle r^2\rangle}$ is equal to the Slater radius of the orbital under question? $\endgroup$ – user157879 Dec 22 '19 at 0:18
  • $\begingroup$ @user157879 I'm not sure I understand your question but maybe my updated answer addresses it. $\endgroup$ – Buck Thorn Dec 22 '19 at 11:09
  • $\begingroup$ Ok I see, so the effective Bohr radius is what zi thought originally, with the addition of an effective $n^*$. Is it just a coincidence for the 3rd shell that $n^*=3$? Do you have a non-paywall link to a table of $n^*$? $\endgroup$ – user157879 Dec 22 '19 at 12:37
  • $\begingroup$ @user157879 You can probably get Slaters original article from another source if you search long enough, since its pretty foundational. I would start with the wikipedia and links therein: en.wikipedia.org/wiki/Slater%27s_rules $\endgroup$ – Buck Thorn Dec 22 '19 at 15:03

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